Answer:
The value is [tex] |P| = 187.4 \ N[/tex]
Explanation:
From the question we are told that
The length of the first rope is L = 6 m
The first rope lie on the x-axis
The end point of the first rope is A
Now the vector of A will be [tex]\vec {OA}= 6i[/tex]
The point of the second rope on the wall is B
The coordinate for the point of the second rope on the wall is (0,-1,2)
The vector of B will be [tex]\vec {OB} = -j+2k[/tex]
Now the coordinate of rope AB is mathematically represented as
[tex]\vec {AB} = \vec{OB} -\vec{OA}[/tex]
=> [tex]\vec {AB} = -j+2k -6i[/tex]
=> [tex]\vec {AB} = -6i -j+2k [/tex]
Generally the magnitude of the rope AB is mathematically evaluated as
[tex]|\vec{AB}| = \sqrt{(-6^2) +(-1)^2 + (2)^2}[/tex]
[tex]|\vec{AB}| = \sqrt{41} \ m [/tex]
Generally the unit vector rope AB is mathematically evaluated as
[tex]\vec r = \frac{\vec {AB}}{|\vec{AB}|}[/tex]
=> [tex]\vec r = \frac{1}{\sqrt{41} } * [-6i -j+ 2k][/tex]
From the question we are told that there is a force acting at point A and the force is
[tex]F = 400 i - 200 j + 500 k[/tex]
Generally the projected component of this force (in N) acting along the rope AB is mathematically represented as
[tex] P = \vec F \cdot \vec r[/tex]
=> [tex] P = 400 i - 200 j + 500 k \ * \ \frac{1}{\sqrt{41} } * [-6i -j+ 2k][/tex]
Note ( i . i = 1 ) , (j . j = 1) , (k . k = 1)
So [tex]P = \frac{1}{\sqrt{41} } [-400 * 6 + 200 * 1 + 500 *2][/tex]
=> [tex]P = -187.4 \ N[/tex]
So the magnitude of the projected component of this force (in N) acting along the rope AB is
[tex] |P| = 187.4 \ N[/tex]
